Question

use the stones u o co n either se the Stonord olmal distribution of the l ist buton to construct a 95 % confidence interval for the population mpon. Justify your decision. If neither distribution con I be used explain why interpret the results. In a random sample of 43 people, the mean body mass index (BMI) WAS 28.5 and the standard deviation I was 6.19 which distribution should be used to construct I the confidence interual? Choose the collect answer brows Al use a normal distribution because the comple is rondom nz 30. and o is known. 2 B. use a normal distribution because the sample is random the population is norrol, and is known. Cluse a t-distribution because the sample is ramdom, nz solt and o is unknown o use a to distribution because the sample is rondom, the - Population is normal and o is unknown E Neither a normal distribution nor a t-distribution con be Lused because either the sample is not randon, olne 30. and the Popniation is not known to be normal I Select the correct choice below and, if necessay, tio in any boxes to complete your choice. Althe 95 90 confidence interval is (D, D CRound to two derimal Place By Neither distribution can be used to construct the confidence Linterual. Interpret the results. Choose the correct answer below. ADIE a 10198 Sandle of people or foten opproximado 95% of them will have a BMI between the bounds of the confidence intool. 2 with 95% confidence, it can be said that the population impon BML is betlupen the bounds of the confidence interual. c It can be between the said that bounds 95 of % the of people have a BMI. confidence interval can be used to construct the D. Neither distribution I Confidence interval.

Answer 1

t distribution is used

option (d)

is TULUI, pluse a to distribution because population is normal and o is the sample unknown

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Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   42          
't value='   tα/2=   2.0181   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   6.1900   / √   43   =   0.943966
margin of error , E=t*SE =   2.0181   *   0.94397   =   1.91
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    28.50   -   1.905001   =   26.594999
Interval Upper Limit = x̅ + E =    28.50   -   1.905001   =   30.405001
95%   confidence interval is (   26.59   < µ <   30.41   )

8 with 95% is between confidence, it can be the bounds of the said that the population meon 8ml confidence interual.

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